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Here is an example, continued fraction for $\sqrt {29}.$ The top row is the "digits," often written $a_i.$ The next row is the convergents, including the two initial fake convergents, $0/1$ and $1/0,$ that begin the process. You can see how the numerator and denominator of a new convergent are specified by the previous two convergents and the "digit." The ...


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With $a(n)=(z+n-1)^2-1$ and $b(n)=z+n,$ the first two identities are $$z=1+\frac{a(1)}{b(1)}, \\ z=1+\frac{a(1)}{1+\frac{a(2)}{b(2)}}.$$ For the continued fraction convergents, the final denominator of $b(n)$ is replaced by $1,$ and so to check convergence we look at the differences $f(n)=C(n)-z$ where $C(n)$ is the $n$th convergent. I found $$f(1)=z^2-z,\\ ...


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Ok, I guess it is by far simple answer, sorry for posting silly questions :(. $\frac{q_n}{q_{n-1}}$ = $\frac{a_n*q_{n-1}+q_{n-2}}{a_{n-1}*q_{n-2}+q_{n-3}}$ = $\frac{a_n*q_{n-1}}{a_{n-1}*q_{n-2}+q_{n-3}}$ + $\frac{q_{n-2}}{a_{n-1}*q_{n-2}+q_{n-3}}$ = $\frac{a_n*q_{n-1}}{q_{n-1}}$ + $\frac{q_{n-2}}{q_{n-1}}$ = $a_n$ + $\frac{q_{n-2}}{q_{n-1}}$ = $a_n$ + $\...


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Ah, at least one empirical pattern which can be built recursively (if that heuristic holds...) Let $$a(n) = \sum_{k=0}^n \frac 1{2^{2^k}} $$ and let $ c(n) $ be the list of partial denominators of the continued fraction (="entries" of the cf) of $a(n)$ - written as a string because we have only few numbers and can nicely compact the notation to focus on the ...


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Shallit's paper, which you've cited, gives a simple algorithm for generating these coefficients. It works for $\sum_{k=0}^{\infty}u^{-2^k}$ for integer $u\ge 3$; but he notes that it also works for $u=2$ (your case) with a slight modification. Start with $$B_1=[1,3].$$ Then repeatedly apply the following rule: $B_{n+1}$ is generated from $B_n$ by appending ...


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Since the ratio test can be applied, we know there exists the limit L. Now we can say $a_n=a_{n+1}$ for $n\to\infty$. If you are just looking for a close approximation, take wolfram alpha https://www.wolframalpha.com/input/?i=sum+%281%2F2^2^k%29 Note that after k=3 nothing dramatic happens anymore, since the fraction gets so small very quickly I don't ...


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Since: $$\mathcal{L}\left(\frac{1}{1+e^x}\right) = \frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right],\qquad \mathcal{L}^{-1}\left(\frac{1}{a+x}\right)=e^{-as}$$ the original integral equals: $$ \int_{0}^{+\infty}\frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds=\frac{1}{2}\...



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